prove that?

Question

|sinh y|<= |cos Z| <= Cosh y

Answer 1

First express cos(z) in terms of the real (x) and imaginary (y) components of z using the addition formula for the cosine:

cos(z) = cos(x+iy)
= cos(x)cos(iy) - sin(x)sin(iy)
= cos(x)cosh(y) - i sin(x)sinh(y)

Then take the absolute square of both sides:

|cos(z)|^2 = cos^2(x)cosh^2(y) + sin^2(x)sinh^2(y)

Obtain the leftmost inequality by replacing sin^2(x) = 1-cos^2(x) in the previous expression:

|cos(z)|^2
=cos^2(x)cosh^2(y) + (1-cos^2(x))sinh^2(y)
=cos^2(x)(cosh^2(y)-sinh^2(y)) + sinh^2(y)
=cos^2(x) + sinh^2(y)
-->
|cos(z)| >= |sinh(y)|

Using the identity cosh^2(y)-sinh^2(y)=1. Similarly, the other inequality follows by replacing cos^2(x) = 1-sin^2(x):

|cos(z)|^2
=(1-sin^2(x))cosh^2(y) + sin^2(x)sinh^2(y)
=cosh^2(y) + sin^2(x)(sinh^2(y)-cosh^2(y))
=cosh^2(y) - sin^2(x)
-->
|cos(z)|^2 <= |cosh(y)| = cosh(y)

Answer 2

o no!
i must be stupid!
thanks for the points anyway!